# $\int_{A-x} f(y) \lambda(dy) = \int_A f(y-x) \lambda(dy)$

Let $$f: \mathbb{R}^d \to \mathbb{R}$$ be measurable and Lebesgue-integrable. Let $$A \in \mathcal{B}(\mathbb{R}^d)$$ a Borel set:

I want to show that:

$$\int_{A-x} f(y) \lambda(dy) = \int_A f(y-x) \lambda(dy)$$

Here is my attempt:

Let $$f = I_B$$ where $$B$$ is Borel set be an indicator function. Then

$$\int_{A-x} f(y) \lambda(dy) = \int_{\mathbb{R}^d} I_{B \cap (A-x)}(y) \lambda(dy)= \lambda(B\cap (A-x))$$

$$\int_A f(y-x) \lambda(dy) = \int_{\mathbb{R}^d}I_{A \cap B}(y-x) \lambda(dy) = \lambda((A\cap B)+x)$$

Are these two equal? I know that $$\lambda(C +x) = \lambda(C)$$.

If I showed it for indicator functions, it also holds for sums of indicatorfunctions. By monotone convergence theorem, it is also true for positive functions and then we can lift this to arbitrary integrable functions by writing such a function as the difference of two positive functions.

When $$f=I_B$$ we have $$\int_A f(y-x)\lambda (dy)=\int_{\mathbb R^{d}} I_A(y)I_B(y-x)\lambda (dy)$$, and not what you have written. The value of this integral is $$\lambda ( A\cap (x+B))$$ because $$y-x \in B$$ iff $$y \in x+B$$. But $$\lambda ( A\cap (x+B))=\lambda ((A-x) \cap B)$$ by translation invariance of Lebesgue measure.
So the desired equality holds when $$f=I_B$$, hence for simple functions and then for non-negative measurable functions by Monotone Convergence Theorem. Finally it holds for integrable $$f$$ by considering $$f^{+}$$ and $$f^{-}$$.